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Chapter 9: Problem 78

Series of squares Prove that if \(\sum a_{k}\) is a convergent series of positive terms, then the series \(\Sigma a_{k}^{2}\) also converges.

Short Answer

Expert verified
Answer: Yes, if the series 鈭� ak converges, then the series 鈭� ak^2 also converges.

Step by step solution

01

Understand the given condition

The given condition is that the series \(\sum a_{k}\) converges, which means that the sum of an infinite number of terms of the series has a finite limit.
02

Comparison Test

To apply the Comparison Test, we need to show that the terms of the squared series \(\sum a_{k}^2\) are smaller (or equal) than the terms of the given convergent series \(\sum a_{k}\). Since the terms are positive, we can compare the ratio of consecutive terms: For \(a_k > 0\), let's consider the ratio of the consecutive terms of the series: \(\frac{a_{k}}{a_{k-1}}>1.\) Raise both sides of the inequality to the power of 2: \(\left(\frac{a_{k}}{a_{k-1}}\right)^2>1 \Rightarrow \frac{a_{k}^2}{a_{k-1}^2}>1.\) Now, multiply both sides of the inequality by \(a_{k-1}^2\) to obtain: \(a_k^2 > a_{k-1}^2.\) This means that the terms of the squared series are decreasing.
03

Monotonicity and Boundedness of Partial Sums of the Squared Series

Since the terms of the squared series are decreasing, we can say that partial sums of the squared series: \(S_n=a_1^2+a_2^2+\cdots+a_n^2\) are both monotonic (non-decreasing) and bounded from above (by the limit of the given convergent series). Therefore, by the Monotone Convergence Theorem, the sequence of partial sums converges, and thus the series \(\sum a_{k}^{2}\) converges.
04

Conclusion

Based on the Comparison Test, we have shown that if the given series \(\sum a_{k}\) converges, then the squared series \(\sum a_{k}^{2}\) also converges.

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Most popular questions from this chapter

Suppose an alternating series \(\sum(-1)^{k} a_{k}\) converges to \(S\) and the sum of the first \(n\) terms of the series is \(S_{n}\) Suppose also that the difference between the magnitudes of consecutive terms decreases with \(k\). It can be shown that for \(n \geq 1,\) $$\left|S-\left[S_{n}+\frac{(-1)^{n+1} a_{n+1}}{2}\right]\right| \leq \frac{1}{2}\left|a_{n+1}-a_{n+2}\right|$$ a. Interpret this inequality and explain why it gives a better approximation to \(S\) than simply using \(S_{n}\) to approximate \(S\). b. For the following series, determine how many terms of the series are needed to approximate its exact value with an error less than \(10^{-6}\) using both \(S_{n}\) and the method explained in part (a). (i) \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\) (ii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k \ln k}\) (iii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}\)

Consider the following infinite series. a. Write out the first four terms of the sequence of partial sums. b. Estimate the limit of \(\left\\{S_{n}\right\\}\) or state that it does not exist. $$\sum_{k=1}^{\infty} \cos (\pi k)$$

Use Exercise 89 to determine how many terms of each series are needed so that the partial sum is within \(10^{-6}\) of the value of the series (that is, to ensure \(R_{n}<10^{-6}\) ). a. \(\sum_{k=0}^{\infty} 0.6^{k}\) b. \(\sum_{k=0}^{\infty} 0.15^{k}\)

A glimpse ahead to power series Use the Ratio Test to determine the values of \(x \geq 0\) for which each series converges. $$\sum_{k=0}^{\infty} x^{k}$$

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty}(-1)^{k} e^{-k}$$
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